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2 years ago

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The weight of a honeybee is 1.2\cdot10^{-1}\text{ g}1.2⋅10 −1 g1, point, 2, dot, 10, start superscript, minus, 1, end superscrip

t, start text, space, g, end text. The weight of the pollen collected by the bee on one trip is 6.0 \cdot 10^{-2} \text{ g}6.0⋅10 −2 g6, point, 0, dot, 10, start superscript, minus, 2, end superscript, start text, space, g, end text.

1 answer:

Nadusha1986 [10]2 years ago

8 0

Answer:1.8*10^-1

Step-by-step explanation:

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Solve the following multiplication and division problems. a. 8 T. 1,398 lb. 14 oz. × 6 b. 349 lb. 6 oz. ÷ 130 c. 6 T. 294 lb. ÷

tester [92]

Answer: a) 278382 oz

b) 43 oz

c) 64.1 oz

Step-by-step explanation:

a) 8 T. 1,398 lb. 14 oz. × 6

First we need to change it in 'oz'.

As we know that

$1\ ton=32000\ oz\\\\1\ lb=16\ oz$

so, it becomes,

$8\times 32000+1398\times 16+14\ oz\\\\=278382\ oz$

8 T. 1,398 lb. 14 oz. × 6 becomes

$278382\times 6\\\\=1670292\ oz$

b) 349 lb. 6 oz. ÷ 130

It becomes,

$349\times 16+6\ oz\\\\=5590\ oz\\\\\text{ at last it becomes}\\\\=\frac{5590}{130}\\\\=43\ oz$

c) 6 T. 294 lb. ÷ 3,071

First it becomes,

$6\times 32000+294\times 16\ oz\\\\=196704\ oz$

At last it becomes,

$\frac{196704}{3071}\\\\=64.1\ oz$

Hence, a) 278382 oz

b) 43 oz

c) 64.1 oz

6 0

2 years ago

Read 2 more answers

On sunday 1/16 people who visit were senior citizens, 3/16 were infants, 3/8 were children and 3/8 were adults, how many of each

mamaluj [8]

Answer:

Out of total people visited the zoo on Sunday , 6.25% was senior citizen , 18.75% were infants , 35% were children and 35% were adult.

Step-by-step explanation:

let total number of people visited the group = x

so number of senior citizens = x × 1/16 = x/16

Percentage of senior citizens = (number of senior citizens ÷ total number of people visited the group) × 100

= (x/16 ÷ x) × 100 = 100/16 = 6.25%

number of infants = 3/16 × x = 3x/16

Percentage of infants =(number of infants ÷ total number of people visited the group) × 100 = (3x/16 ÷ x) × 100 = 300/16 = 18.75%

number of children = 3/8 × x = 3x/8

Percentage of children =(number of children ÷ total number of people visited the group) × 100 = (3x/8 ÷ x) × 100 = 300/8 = 37.5%

number of adults = 3/8 × x = 3x/8

Percentage of adults =(number of adults ÷ total number of people visited the group) × 100 = (3x/8 ÷ x) × 100 = 300/8 = 37.5%.

Hence we can conclude that Out of total people visited the zoo on Sunday , 6.25% was senior citizen , 18.75% were infants , 35% were children and 35% were adult.

4 0

2 years ago

Astrid is in charge of building a new fleet of ships. Each ship requires 404040 tons of wood, and accommodates 300300300 sailors

prisoha [69]

Answer:

10 Ships

Step-by-step explanation:

Source: Khan Academy

4 0

2 years ago

Read 2 more answers

World poultry production was 77.2 million tons in the year 2004 and increasing at a continuous rate of 1.6% per year. Assume tha

dezoksy [38]

Answer:

A) $P(t)=77.2\cdot e^{0.016t}$

B) 89.157 million tons.

C) In year 2017.

Step-by-step explanation:

We have been given that world poultry production was 77.2 million tons in the year 2004 and increasing at a continuous rate of 1.6% per year.

A). We know that a continuous growth function is in form $y=a\cdot e^{kx}$ , where,

a = Initial value,

k = Growth rate in decimal form.

$1.6\%=\frac{1.6}{100}=0.016$

Upon substituting our given values, we will get:

$P(t)=77.2\cdot e^{0.016t}$

Therefore, our required function would be $P(t)=77.2\cdot e^{0.016t}$ , where, P represents world poultry production, in million tons, as a function of the number of years, t, since 2004.

B) To find the world poultry production in the year 2013, we will substitute $t=2013-2004=9$ in above formula as:

$P(9)=77.2\cdot e^{0.016(9)}$

$P(9)=77.2\cdot e^{0.144}$

$P(9)=77.2\cdot 1.1548841085249135$

$P(9)=89.1570531781\approx 89.157$

Therefore, the world poultry production in the year 2013 would be 89.157 million tons.

C) To find the number of years it will take for world poultry production to be over 95 million tons, we will equate our function with 95 as:

$95=77.2\cdot e^{0.016t}$

Divide both sides by 77.2:

$1.2305699481865285=e^{0.016t}$

Take natural log of both sides:

$\text{ln}(1.2305699481865285)=\text{ln}(e^{0.016t})$

$\text{ln}(1.2305699481865285)=0.016t\text{ln}(e)$

$0.20747743456981035=0.016t*1$

Divide both sides by 0.016:

$t=12.9673396606$

$t\approx 13$

Therefore, 13 years after in 2017 world poultry production goes over 95 million tons.

4 0

2 years ago

A sanitation supervisor is interested in testing to see if the mean amount of garbage per bin is different from 50. In a random

Harlamova29_29 [7]

Answer with explanation:

Given : Sample mean = $\overline{x}=\text{48.47 pounds}$

Standard deviation : $\sigma=\text{ 3.1 pounds.}$

Sample size : n = 36

Claim : $\mu\neq50$

∴ $H_0:\mu=50$

$H_1:\mu\neq50$

Since the alternative hypothesis is two tail , then the test is two tail test.

By using a z statistic and a 0.05 level of significance. Reject $H_0$ if z < -1.960 or is z> 1.960.

Then , the test static for population mean is given by :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

$z=\dfrac{48.47-50}{\dfrac{3.1}{\sqrt{36}}}\approx-2.96$

We reject $H_0$ since $-2.96\leq -1.96$ . We have statistically significant evidence at $\alpha=0.05$ to show that the mean amount of garbage per bin is not different from 50.

5 0

2 years ago